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Questions tagged [cardinals]

This tag is for questions about cardinals and related topics such as cardinal arithmetics, regular cardinals and cofinality. Do not confuse with [large-cardinals] which is a technical concept about strong axioms of infinity.

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Irrational numbers Cardinality.

The real numbers, $\mathbb{R}$, are uncountable and the rational numbers, $\mathbb{Q}$, are countable. We can write $\mathbb{R} = \mathbb{Q} \cup (\mathbb{R} \setminus \mathbb{Q})$. Since $\mathbb{Q}$ ...
Mathstudent123's user avatar
-1 votes
3 answers
89 views

Find the cardinality of $A \cup B$ [closed]

Let the sets $A=\{\frac{1}{1\times 2} , \frac{1}{2\times 3}, ... , \frac{1}{2021\times 2022}\}, B=\{ \frac{1}{2\times 4}, \frac{1}{3\times 5}, ..., \frac{1}{2020\times 2022}\}$. Find the cardinality ...
Pam Munoz Ryan's user avatar
0 votes
1 answer
54 views

For $k$-algebras $B_1, \dots, B_n$, $\# \operatorname{Hom}_k( \prod_{i=1}^nB_i, \Omega) = \Sigma_{i=1}^n \# \operatorname{Hom}_k(B_i, \Omega)$?

Let $k$ be a field with $ k \subseteq \Omega$ a algebraically closed field. Let $B_1 , \dots, B_n$ be ( possibly finite local ) $k$-algebras. Then next equality of cardinals holds $$ \# \operatorname{...
Plantation's user avatar
  • 2,718
0 votes
1 answer
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Proving the Equality of Infinite Cardinal Products and Powers

Theorem: Let $\Xi$ be an infinite set, $\{\kappa_i\}_{i \in \Xi}$ be a family of cardinal numbers, and $\lambda$ be a cardinal number. Then: $\prod_{i \in \Xi} \kappa_i^{\lambda} = \left(\prod_{i \in \...
Chau Long's user avatar
5 votes
1 answer
114 views

Absoluteness of inaccessible cardinals

I'm studying large cardinals and I'm hoping to fully understand the proof that says ZFC is not able to prove the existence of inaccessibles (given ZFC is consistent, of course). I've already fully ...
Darsen's user avatar
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1 vote
1 answer
59 views

Brun's theorem and the twin prime conjecture

According to the following extract taken from Wikipedia, almost all prime numbers are isolated given Brun's theorem. Doesn't that mean that there is only a finite number of twin prime numbers (they ...
David's user avatar
  • 37
1 vote
1 answer
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Cardinal of a set of integers with symmetry relations

Context In computational chemistry, there are two-electron integrals noted $(ij|kl)$ for integers (i,j,k,l) between 1 and K. The explicit expression of $(ij|kl)=\int dx_1dx_2 \chi_i(x_1)\chi_j(x_1)\...
L Maxime's user avatar
2 votes
0 answers
43 views

Lemma 0 in Hajnal's Paper "Embedding Finite Graphs into Graphs Colored with Infinitely Many Colors"

I am looking for a proof of the following lemma. Let $E_0$ be the set of edges of an undirected graph with no loops with vertex set a cardinal $\kappa$. Let $E_1$ be the family of two-element subsets ...
Tri's user avatar
  • 417
2 votes
1 answer
70 views

Assuming GCH holds, calculate $\aleph_{\omega_1}^{\aleph_0}$

I'm working through the book Discovering Modern Set Theory by Just and Weese, and this question comes right after this theorem: Here's what I've worked out so far: I believe the cofinality of $\...
violeta's user avatar
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4 votes
2 answers
168 views

Does the Cardinal Supremum Commute with the Cardinal Power?

Let $\kappa,\tau$ be two cardinals and $\{\varkappa_\alpha\}_{\alpha<\kappa}$ an indexed set of cardinals. Is it true that $$\sup_{\alpha<\kappa}(\varkappa_\alpha^\tau)=\left(\sup_{\alpha<\...
tripaloski's user avatar
0 votes
1 answer
85 views

For infinite cardinals $\kappa$, we have $\kappa \otimes \kappa = \kappa$.

I am aware that other questions are quite similar to this; however, it seems like the other questions regarding the same statement are looking at proofs that seem somewhat different from the one I am ...
Ben123's user avatar
  • 1,309
0 votes
2 answers
58 views

Cardinality of a set of disjoint open sub intervals of $( 0 ,1)$

Let $A$ be any collection of disjoint open subintervals of $(0 ,1)$ . Then what is maximum cardinality of $A$ ? I know one easy way to prove its countable is that every open interval has rational ...
User492177's user avatar
1 vote
0 answers
29 views

A sequence of continuum hypotheses

The continuum hypothesis asserts that $\aleph_{1}=\beth_{1}$. Both it and its negation can be consistent with ZFC, if ZFC is consistent itself. The generalised continuum hypothesis asserts that $\...
Darmani V's user avatar
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1 answer
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Can cardinality $\kappa$ exist where $\forall n\in\mathbb{N} \beth_n<\kappa$,$\kappa<|\bigcup_{n\in\mathbb{N}}\mathbb{S}_n|$,$|\mathbb{S}_n|=\beth_n$

The Wikipedia article on Beth numbers defines $\beth_\alpha$ such that $\beth_{\alpha} =\begin{cases} |\mathbb{N}| & \text{if } \alpha=0 \\ 2^{\beth_{\alpha-1}} & \text{if } \alpha \text{ is a ...
SarcasticSully's user avatar
1 vote
2 answers
142 views

proving the set of natural numbers is infinite (Tao Ex 2.6.3)

Tao's Analysis I 4th ed has the following exercise 3.6.3: Let $n$ be a natural number, and let $f:\{i \in \mathbb{N}:i \leq i \leq n\} \to \mathbb{N}$ be a function. Show that there exists a natural ...
Penelope's user avatar
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1 vote
0 answers
55 views

Injective monotonic mapping from rationals $\mathbb Q^2$ to $\mathbb R$

Exercise: $f: \mathbb Q^2\to\mathbb R$. Where $\mathbb Q$ is the set of rational numbers. $f$ is strictly increasing in both arguments. Can $f$ be one-to-one? This question is related to many ...
High GPA's user avatar
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1 vote
0 answers
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Can a countable union of subgroups of uncountable index in G be equal to G? [closed]

Let G be a group and $\{H_i\}_{i<\omega}$ be a countable family of subgroups of $G$, each of them of uncountable index. Can $G=\bigcup_{i<\omega} H_i$?
Moreno Invitti's user avatar
2 votes
4 answers
232 views

Is the cardinality of $\varnothing$ undefined?

It is intuitive that the cardinality of the empty set is $0$. However we are asked to demonstrate this using given definitions/axioms in Tao Analysis I 4th ed ex 3.6.2. My question arises as I think ...
Penelope's user avatar
  • 3,325
0 votes
0 answers
43 views

How does measure theory deal with higher cardinalities?

The second part of the definition of a sigma-algebra is that countable unions of measurable sets are measurable. The second property of a measure is that the measure of countable unions of measurable ...
TheOwl's user avatar
  • 11
0 votes
0 answers
56 views

On the Singular Cardinal Hypothesis

I'm trying to find the proof of this result. If for each $\lambda\geq2^\omega$, $\lambda^\omega\le\lambda^+$, then the SCH holds. I'm not sure where to look. So if you have any info about this, please ...
Selena's user avatar
  • 103
0 votes
1 answer
66 views

Why is $\{0,1\}^{\Bbb N}$ uncountable? [duplicate]

In the book Measure and Integral : An introduction to real analysis, in chapter 8 Lp spaces, theorem 8.18, the authors give a counterexample to show that $l^\infty$ is not separable: Question: why is ...
G.t.g.h's user avatar
  • 161
0 votes
0 answers
29 views

Cardinal sum of powers

I'm trying to solve this exercise. Can anybody please help me: If $\kappa$ ist an infinite cardinal number with $cf(\kappa) = \kappa$ and for all $\mu < \kappa$ the inequality $2^{\mu} \leq \kappa$ ...
Johannes's user avatar
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0 answers
34 views

For all cardinals $\kappa, \lambda$ with $\lambda \geq cf(\kappa)$ the inequality $\kappa^{\lambda} > \kappa$ holds [duplicate]

I genuinely have no idea why the proposition in the title holds or how to show it. I am kind of new to cardinals and ordinals and very confused. If someone could explain, I would really appreciate ...
metamathics's user avatar
1 vote
0 answers
92 views

Cardinal power towers

I am not an expert on large cardinals. I could not find any reference (and terminology) for the following question: We start with $$\lambda:=\aleph_0 \text{ [tet] } \omega = \aleph_0 ^ {\aleph_0 ^ {\...
Cem Aksak's user avatar
0 votes
0 answers
25 views

Would well-founded Scott cardinals work in ZCA + Ranks?

Does original Zermelo's set theory + Regularity + Ranks, prove that every set is of equal size to some element of a Scott cardinal? The original Zermelo does include an axiom of Choice, and it admits ...
Zuhair's user avatar
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1 vote
1 answer
119 views

The cardinality of specific set $A\subset \mathbb{N}^{\mathbb{N}}$

Let $A$ be a set of total functions from the naturals to the naturals  such that for every $f\in A$ there is a finite set $B_f\subset \mathbb{N}$ , such that for every $x\notin B_f$ , $f(x+1)=f(x)+1$. ...
Daniel's user avatar
  • 83
4 votes
1 answer
82 views

Induction does not preserve ordering between cardinality of sets?

Consider building a binary tree and consider it as a collection of points and edges. Here is one with five levels, numbered level $1$ at the top with $1$ node to level $5$ at the bottom with $16$ ...
jdods's user avatar
  • 6,360
3 votes
1 answer
107 views

A family of $\kappa^{<\omega}$ such that for every member in $\kappa$ is contained by all but finitely many elements of the family

Suppose that $\kappa$ is an uncountable cardinal. Let $\kappa^{<\omega}$ denote the family of all finite subsets of $\kappa$. Does there exist a family $S\subset\kappa^{<\omega}$ such that for ...
Jianing Song's user avatar
  • 1,923
0 votes
0 answers
55 views

How should I should prove $\mathbb{R}\sim\{0,1\}^{ \mathbb{N}}$ [duplicate]

I've seen some argument about the binary representation, but I think it is not accurate because under some extreme cases, the rounding or bit constraint would results distinct reals also have the same ...
LJNG's user avatar
  • 1,364
2 votes
1 answer
309 views

Dependence of the equation $a+1=a$ for infinite cardinals $a$ on the axiom of choice

let $A$ be a set such that for all $n \in $ N $ A ≉ N_n$ where $N_n = \{ 0 ,1 ,2 ...... n-1\} $ and $a$ be the Cardinality of $A$ meaning ($|A| = a$) is it possible to prove that $a+1=a$ without ...
theorem 28's user avatar
2 votes
1 answer
75 views

The cardinality of a convergent series [duplicate]

$A\subseteq \mathbb{R}^{+}$  is a set of positive real numbers ($0\notin \mathbb{R}^{+}$), for which there exists a positive real number $x$, such that for every finite subset $S\subseteq A$, the sum ...
Daniel's user avatar
  • 83
2 votes
1 answer
44 views

Singular cardinals and $\kappa$-Lindelöf spaces

Say a space is $\kappa$-Lindelöf provided that for every open cover of the space, there exists a subcover of cardinality $<\kappa$. So $\aleph_0$-Lindelöf is compact, and $\aleph_1$-Lindelöf is ...
Steven Clontz's user avatar
0 votes
2 answers
57 views

Are there any theorems that use the uncountability of the reals in their proof?

Can we use the uncountability of the reals as a tool to prove any theorem? Can we use this to calculate anything? Suppose I was trying to convince a pragmatist that uncountability is useful.
Alex's user avatar
  • 470
0 votes
0 answers
31 views

Prove $C ∼ P(P(\mathbb{N}))$ when $C$ is defined as the set of all $S$ s.t $(z − m, z + m) ∩ S = ∅ $ for every $z∈\mathbb{Z}$, $m∈\mathbb{R}$

First, I know that there is a very similar question - The cardinality of all sets $A$ such that $\forall \ z\in \mathbb{Z} \ , (z-k,z+k)\ \cap A=\emptyset : 0<k<0.5$ , but here I want to the ...
User33975329257439645's user avatar
1 vote
1 answer
117 views

The cardinality of all sets $A$ such that $\forall \ z\in \mathbb{Z} \ , (z-k,z+k)\ \cap A=\emptyset : 0<k<0.5$

The question asks the following: Given $k\in \mathbb{R}$ such that $0<k<0.5$ a set $A$ is "k-integer-avoidant" if $ \ \forall z\in\mathbb{Z}$ , $\left ( z-k,z+k \right ) \cap A=\...
Daniel's user avatar
  • 83
1 vote
3 answers
92 views

$B\subseteq P(\mathbb{N}) : |B| = \aleph \ ?$

The question asked if there exist a set $B\subseteq P(\mathbb{N})$ with cardinality of $\aleph$ such that for all $A_1 ,A_2 \in B$ , if $A_1\neq A_2$ then $A_1 \cap A_2 = \phi$ . I have looked at the ...
Daniel's user avatar
  • 83
2 votes
1 answer
47 views

Saturation of infinite complete Boolean algebra is a regular cardinal

Similar question existed here. However there are still many gaps for stupid persons like me. A Boolean algebra $B$ is called $\kappa$-saturated if there is no antichain with supremum $1$ (also called ...
BlowingWind's user avatar
0 votes
0 answers
59 views

Finite cardinals raised to the power of an infinite cardinal

I am trying to prove the fact that if $a$ and $b$ are finite cardinals, and $c$ is an infinite cardinal, then $a^c = b^c$. I am able to prove this fact by using $d \cdot d = d$ for all infinite ...
Mark Worrall's user avatar
0 votes
0 answers
46 views

Prove that an infinite well ordered set X has equal cardinality to the set X∪{a}, where 'a' does not belong to X.

Found this question in a book of analysis as a corollary. Before the question is introduced (as an exercise), the book introduced Theorem of Recursion on Wosets and Comparability Theorem. For ...
Arbbor1's user avatar
1 vote
0 answers
42 views

What is the cardinality of the Diffeomorphism group $\text{Diff}(\mathbb{R})$ over the reals?

As a small set-up, we have $\aleph_0$ as the cardinality of $\mathbb{N}$, and then the cardinality of the reals $\mathbb{R}$ is $\beth_1 = 2^{\aleph_0}$. In particular, it turns out that this is also ...
Yannik Wotte's user avatar
0 votes
2 answers
63 views

Prove cardinality of R2 without creating a map to R1

I think we can prove $|\mathbb{R}^2|=\aleph_1$, by creating a bijection between $\mathbb{R}$ and $\mathbb{R}^2$. But this map is difficult to construct. Is there any easier way to show $|\mathbb{R}^2|=...
Alex's user avatar
  • 470
5 votes
2 answers
124 views

Why does countability misbehave in intuitionistic logic

On page 3 of this paper https://arxiv.org/pdf/2404.01256.pdf I spotted the claim: Definitions of countability in terms of injection into ℕ misbehave intuitionistically, because a subset of a ...
Y.X.'s user avatar
  • 4,223
0 votes
1 answer
88 views

Is there a way to construct larger cardinals without choice axiom?

From Cantor's Theorem, we know that $|\mathcal{P}(X)| > |X|$. So, we can define inductively a set with cardinality $\aleph_n, \forall n \in \mathbb{N}$. Let $\lbrace A_i\rbrace_{i \in \mathbb{N}}$ ...
Edwin's user avatar
  • 31
0 votes
0 answers
53 views

Is this Proof Involving Cardinal Arithmetic Correct?

Question Show that for $n>0$, $n.2^{2^{\aleph _0}}=\aleph _0.2^{2^{\aleph _0}}= 2^{\aleph _0}.2^{2^{\aleph _0}}= 2^{2^{\aleph _0}}.2^{2^{\aleph _0}}=(2^{2^{\aleph _0}})^n=(2^{2^{\aleph _0}})^{\...
Mr Prof's user avatar
  • 451
4 votes
1 answer
105 views

Among 101 dalmatian dogs, each dog has a unique number of black spots, Addition property

Among 101 dalmatian dogs, each dog has a unique number of black spots from the set {1, 2, 3, . . . , 101}. We choose any 52 of the 101 dogs. We want to prove that any set of 52 dogs satisfies the ...
Asher's user avatar
  • 343
17 votes
2 answers
403 views

Combinatorial proof, without axiom of choice, that for any set $A$, there is no surjection from $A^2$ to $3^A$

The well known proof of Cantor's theorem (stating that $A<2^A$ for any set $A$) does not make any use of the axiom of choice. I have now spent some time wondering if the analogous result $A^2<3^...
Tim Seifert's user avatar
  • 2,243
13 votes
2 answers
827 views

Infinite wacky race

Dick Dastardly is taking part in an infinite wacky race. What is infinite about it, you ask? Well, just everything! There are infinitely many racers, every one of which can run infinitely fast and the ...
Alma Arjuna's user avatar
  • 3,901
1 vote
1 answer
186 views

Fixing my gripe with the common proof for showing that $|\mathcal{P}(\mathbb{N})| = |\mathbb{R}|$

I am familiar with the proof that shows the powerset of the naturals is of the same cardinality as the reals using binary representation. Here's a quick rundown of the proof: Showing that $f:(-1, 1) \...
Vector's user avatar
  • 377
2 votes
1 answer
76 views

How can I count the number of eventually constant functions $\kappa_1\to\kappa_2$?

Given two infinite cardinals $\kappa_1,\kappa_2$, what's the number $\tau$ of functions $f:\kappa_1\to\kappa_2$ that are eventually constant? I think that, if $\tau_0$ is the number of eventually zero ...
tripaloski's user avatar
3 votes
1 answer
73 views

Is it consistent with ZC that a well-order of type $\omega_\omega$ does not exist?

Working in Zermelo's set theory (with choice for simplicity) - the construction in Hartogs' theorem shows that starting with a set $X$, there is a set $X'$ in at most $\mathcal{P}^4(X)$ (where $\...
Chad K's user avatar
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